Electric field at center of semi-circular bent rod of charge

[caesius]

I have the correct answer just the wrong direction

Homework Statement

A uniformly charged insulating rod of length 14.0cm is bent into the shape of a semicircle as shown in figure XXX. The rod has a total charge of 7.5 micro C. Find the magnitude and direction of the electric field at P, the center of the semicircle.

Homework Equations

dE = k.dq.1/r^2

The Attempt at a Solution

I set things up like this
http://img210.imageshack.us/img210/4690/physprobub2.png

Since the vertical components would cancel I just looked at the x components.

<br /> dE_{x} = \frac{k_{e}\cdot dq\cdot\sin\theta}{r^2} <br />

<br /> = \frac{k_{e}\cdot dx\cdot\lambda\cdot\sin\theta}{r^2} \ \ \mbox{as dq = dx\cdot\lambda}<br />

<br /> = \frac{k_{e}\cdot r\cdot d\theta\cdot\lambda\cdot\sin\theta}{r^2} \ \ \mbox{as dx = ds = r\cdot d\theta}<br />

<br /> = \frac{k_{e}\cdot d\theta\cdot\lambda\cdot\sin\theta}{r} \ \ \mbox{as dx = ds = r\cdot d\theta}<br />

So
<br /> E = \frac{k_{e}\cdot \lambda}{r} \int d\theta\cdot\sin\theta}<br />
(integral is from zero to pi)

<br /> E = \frac{k_{e}\cdot \lambda}{r} ( -1 - 1)<br />

<br /> E = \frac{-2\cdot k_{e}\cdot Q}{L\cdot r} \ \ \mbox{as \lambda = \frac{Q}{L}}<br />
When I substitute the values is I get the correct answer but it's negative. The bent rod is positive so the resultant electric field should be AWAY from the semicircle and therefor positve?!?

What have I done wrong?

 

[Dick]

The integral of sin from 0 to pi is 2. How did you get (-1-1)? BTW the antiderivative of sin(x) is -cos(x).

 

[IAmBatman07]

Sorry for reactivating this old thread, but I figured better that than starting a whole new thread.

I have gotten to the point of <br /> <br /> E = \frac{2\cdot k_{e}\cdot Q}{L\cdot r} \ <br /> <br />

but now my question is what is the r value? Of what my reasoning is telling me, it should be <br /> <br /> E = \frac{2\cdot k_{e}\cdot Q}{L^{2}} \ <br />

Is this right? Can anyone assist me?